Question

If cos(α – β) + cos(β – γ) + cos(γ – α) = −32, then prove that cos α + cos β + cos γ = sin α + sin β + sin γ​

Answer 1

Answer:

answer is A. B And D

Step-by-step explanation:

Correct option is

A

∑cosα=0

B

∑sinα=0

D

∑(cosα+sinα)=0

Given cos(β−γ)+cos(γ−α)+cos(α−β)=−

2

3

⟹cosβcosγ+sinβsinγ+cosγcosα+sinγsina+cosαcosβ+sinasinβ=−

2

3

⟹2(cosβcosγ+sinβsinγ+cosγcosα+sinγsina+cosαcosβ+sinasinβ)+1+1+2=0

⟹2(cosβcosγ+cosγcosα+cosαcosβ)+2(sinβsinγ+sinγsinα+sinαsinβ)+(sin

2

α+cos

2

α)

+(sin

2

β+cos

2

β)+(sin

2

γ+cos

2

γ)=0

⟹(cosα+cosβ+cosγ)

2

+(sinα+sinβ+sinγ)

2

=0

⟹∑cosα=0 and ∑sinα=0

⟹∑(cosα+sinα)=0

Hence, options 'A', 'B' and 'D' are correct.

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Answer 2

(The question is a little incorrect. It should be -3/2 instead of -32)

cos(α – β) + cos(β – γ) + cos(γ – α) = −3/2

cos α cos β + sin α sin β + cos β cos γ + sin β sin γ + cos γ cos α + sin γ sin α = -3/2

2(cos α cos β + sin α sin β + cos β cos γ + sin β sin γ + cos γ cos α + sin γ sin α) + 1 + 1 + 1 = 0

2(cos α cos β + cos β cos γ + cos γ cos α) + 2(sin α sin β + sin β sin γ + sin γ sin α) + (sin²α+cos²α) + (sin²β+cos²β) + (sin²γ+cos²γ) = 0

{cos²α + cos²β + cos²γ + 2(cos α cos β + cos β cos γ + cos γ cos α) } + {sin²α + sin²β + sin²γ + 2(sin α sin β + sin β sin γ + sin γ sin α) } = 0

(cos α + cos β + cos γ)² + (sin α + sin β + sin γ)² = 0

Therefore, cos α + cos β + cos γ = sin α + sin β + sin γ = 0